1

Chapter 7

Continuous Probability Distributions

2

Goals1. Understand the difference between

discrete and continuous distributions2. Compute the mean and the standard

deviation for a uniform distribution3. Compute probabilities using the uniform

distribution

3

Goals4. List the characteristics of the:

• Normal probability distribution• Standard normal probability distribution

5. Define and calculate z values6. Use the standard normal probability distribution

to find area: Above the mean Below the mean Between two values Above one value Below one value

7. Use the normal distribution to approximate the binomial probability distribution

4

Probability Distribution A listing of all the outcomes of an experiment

and the probability associated with each outcome

Probability distributions are useful for making probability statements concerning the values of a random variable

Our goal is to find probability between two values: Example: What is the probability that the daily water

usage will lie between 15 and 25 gallons? A: 68%

5

Probability Distribution

Discrete Probability Distributions (Chapter 6) Based On Discrete Random Variables We looked at:

Binomial Probability Distribution

Continuous Probability Distribution (Chapter 7) Based On Continuous Random Variables: We will look at:

Uniform Probability Distribution Normal Probability Distribution

6

Continuous Probability Distributions

These continuous probability distributions will be all about

Area!!!!

Possibilities within rangea - b

-3 -2 -1 0 1 2 3 zμ + -3 σ μ + -2 σ μ + -1 σ μ μ +1 σ μ +2 σ μ +3 σ x symbols

1400 1600 1800 2000 2200 2400 2600 x in dollars

68.26% area

95.44% area

99.74% area

34.13% area

13.58% area

2.15% area

0.13% area

34.13% area

13.58% area

2.15% area

0.13% area

7

Continuous Probability Distribution: Uniform Probability Distribution

Within the interval 15 to 25 minutes, the time it takes to fill out a typical 1040EZ tax return at a VITA site tends to follow a uniform distribution Random Variable is time (only possibilities within the interval) Each value has same probability

Normal Probability Distribution The weight distribution of a manufactured box of cereal

tends to follow a normal distribution Random Variable is box weight and will cover all possibilities Each value has different probability

8

Uniform Probability Distribution1. Distributions shape is rectangular2. Minimum value = a3. Maximum value = b

a and b imply a range

4. Height of the distribution is constant (uniform) for all values between a and b

Implies all values in range are equally likely

Minimum Value = aMaximum Value = bMean of Uniform Distribution = (a+b)/2 = µStandard Deviation of Uniform Distribution = (((b-a) 2̂)/12)) (̂1/2)) = sIf a <= x <= b and 0 everywhere else = 1/(b-a) = P(x)If a <= x <= b and 0 everywhere else = 1/(b-a) Height of RectangleArea = Height * Base = (1/(b-a))*(b-a) = 1

9

Uniform Probability DistributionP(x)

1All Posibilities

b-a Possibilities within rangea - b

Unitsa b

< ------- b-a ------- >

2

ba

12

)( 2ab s

elsewhere 0 and if ,

1)( bxa

abxP

00.1)()(

1*

ab

abBaseHeightArea

10

Suppose the time that you wait on the telephone for a live representative of your phone company to discuss your problem with you is uniformly distributed between 5 and 25 minutes.

What is the mean wait time?

a + b 2 =

= 5+25 2

= 15

What is the standard deviation of the wait time?

s =(b-a)2

12

= (25-5)2

12= 5.77

11

What is the probability of waiting more than ten minutes?

The area from 10 to 25 minutes is

15 minutes. Thus:

P(10 < wait time < 25) = height*base = 1

(25-5) *15 = .75

1 = 125-5 20

minutes

P(x

)

5 10 15 20 25

12

What is the probability of waiting between 15 and 20 minutes?

The area from 15 to 20 minutes is

5 minutes. Thus:

P(15 < wait time < 20) = height*base = 1

(25-5)*5 = .25

1 = 125-5 20

minutes

P(x

)

5 10 15 20 25

13

μ + -3 σ μ + -2 σ μ + -1 σ μ μ +1 σ μ +2 σ μ +3 σ x (values) symbols

68.26% area

95.44% area

99.74% area

34.13% area

13.58% area

2.15% area

0.13% area

34.13% area

13.58% area

2.15% area

0.13% area

Normal Probability Distribution Is All About Area!Total Area = 1.0

14

Normal Probability Distribution Formula

2

2

2

)(

2

1)( s

s

x

exP

Awesome!

No, that’s o.k., we can use Appendix or Excel functions!

x = 250s = 2µ = 247

1.5 =(I3-I5)/I4 = 0.0648 0.064759 =NORMDIST(I3,I5,I4,FALSE)

0.933193 =NORMDIST(I3,I5,I4,TRUE)250 =NORMINV(K8,I5,I4)

0.933193 =NORMSDIST((I3-I5)/I4)1.5 =NORMSINV(K10)

2

2

2

)(

2

1)( s

s

x

exP

15

Characteristics of a NormalProbability Distribution

(And Accompanying Normal Curve)

The normal curve is bell-shaped and has a single peak at the exact center of the distribution

The arithmetic mean, median, and mode of the distribution are equal and located at the peak Thus half the area under the curve is above the mean

and half is below it

The normal probability distribution is symmetrical about its mean If we cut the normal curve vertically at this center

value, the two halves will be mirror images

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The normal probability distribution is asymptotic The curve gets closer and closer to the X-axis but

never actually touches it The “tails” of the curve extend indefinitely in both

directions The Location of a normal distribution is

determined by mean µ The dispersion of a normal distribution is

determined by the standard deviation s

Characteristics of a NormalProbability Distribution

(And Accompanying Normal Curve)

Now Let’s Look At Some Pictures That Will ShowRelationships Amongst Various Means & Standard Deviations

17

Characteristics of a Normal DistributionNormalcurve issymmetrical

Theoretically,curveextends toinfinity

Mean, median, andmode are equal

There Is A Family Of Normal Probability Distributions

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Which One Is Normal?

Years

Fre

qu

ency

Years

Fre

qu

ency

Years

Fre

qu

ency

Years

Fre

qu

ency

19

Equal Means, Unequal Standard Deviations

σ = 3.1 years

μ = 20 years Length Of Service

20

σ = 3.1 years

μ = 20 years Length Of Service

σ = 3.9 years

Equal Means, Unequal Standard Deviations

21

σ = 3.1 years

μ = 20 years Length Of Service

σ = 3.9 years

σ = 5.0 years

Equal Means, Unequal Standard Deviations

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Unequal Means, Equal Standard Deviations

σ = 1.6 Grams

μ = 321 Grams of Super Rad Cereal

σ = 1.6 Grams

μ = 301 Grams of Super Neat Cereal

σ = 1.6 Grams

μ = 283 Grams of Super Yummy Cereal

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Unequal Means, Unequal Standard Deviationsσ = 26 psi

σ = 52 psi

μ = 2107 psi

σ = 41 psi

μ = 2000 psi μ = 2186 psi

This Family Of Normal Probability DistributionsIs Unlimited In Number!

Luckily, One Of The Family Members May Be Used In AllCircumstances Where The Normal Distribution Is Applicable

Standard Normal Distribution

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The standard normal distribution (z distribution ) is a normal distribution with a mean of 0 and a standard deviation of 1

The percentage of area between two z-scores in any normal distribution is the same! Standard deviation & terms may be different, but area will

be the same!

Normal distributions can be converted to the standard normal distribution using z-values…

Standard Normal Probability Distribution

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Define And Calculate z-values Any normal distribution can be converted, or “standardized”

to the standard normal distribution using z-values Z-values:

Distance from the mean, measured in units of standard deviation

The Formula Is:

s

X

z

z = z-scorex = particular valueμ = meanσ = standard deviation

Define VariablesZ-values are also called:

Standard normal valueZ scoreZ statisticStandard normal deviateNormal deviate Remember Your Algebra

So That You Can Solve ForAny One Of The Variables

26

-3 -2 -1 0 1 2 3 zμ + -3 σ μ + -2 σ μ + -1 σ μ μ +1 σ μ +2 σ μ +3 σ x symbols

1400 1600 1800 2000 2200 2400 2600 x in dollars

0 means that there is no deviation from the mean!

Standard Normal Probability Distribution

27

Convert Value From A Normal Distribution To A Z-score Example 1

The bi-monthly starting salaries of recent MBA graduates follows the normal distribution with a mean of $2,000 and a standard deviation of $200

What is the z-value for a salary of $2,200?

$2,200 $2,0001.00

$200

Xz

s

28

Convert Value From A Normal DistributionTo A Z-score, Example 2

What is the z-value of $1,700?

$1,700 $2,0001.50

$200

Xz

s

A z-value of 1 indicates that the value of $2,200 is one standard deviation above the mean of $2,000 (2000 + 1*200)

A z-value of –1.50 indicates that $1,700 is 1.5 standard deviation below the mean of $2000 (2000 – 1.5*200)

Now we can look at a graph

29

What is the probability that a foreman’s salary will fall between

1,700 and $2,200?

30

But It Is Really All About Area Under The CurveRemember:

Areas Under the Normal Curve• Empirical Rule (Normal Rule):

• About 68% of the observations will lie within 1 σ of the mean

• About 95% of the observations will lie within 2 σ of the mean

• Virtually all the observations will be within 3 σ of the mean

Hints: Many statistical chores can be solved with this normal curveNevertheless, “The whole world does not fit into a normal curve”

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-3 -2 -1 0 1 2 3 zμ + -3 σ μ + -2 σ μ + -1 σ μ μ +1 σ μ +2 σ μ +3 σ x symbols

1400 1600 1800 2000 2200 2400 2600 x in dollars

68.26% area

95.44% area

99.74% area

34.13% area

13.58% area

2.15% area

0.13% area

34.13% area

13.58% area

2.15% area

0.13% area

Between what two values do about 95% of the values occur?

Empirical Rule (Normal Rule):

What if you want to find the % of values that lie between z-scores 0 and 1.56?

32

z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.090.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239 0.0279 0.0319 0.03590.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.07530.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.0987 0.1026 0.1064 0.1103 0.11410.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.1368 0.1406 0.1443 0.1480 0.15170.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.1736 0.1772 0.1808 0.1844 0.18790.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088 0.2123 0.2157 0.2190 0.22240.6 0.2257 0.2291 0.2324 0.2357 0.2389 0.2422 0.2454 0.2486 0.2517 0.25490.7 0.2580 0.2611 0.2642 0.2673 0.2704 0.2734 0.2764 0.2794 0.2823 0.28520.8 0.2881 0.2910 0.2939 0.2967 0.2995 0.3023 0.3051 0.3078 0.3106 0.31330.9 0.3159 0.3186 0.3212 0.3238 0.3264 0.3289 0.3315 0.3340 0.3365 0.33891.0 0.3413 0.3438 0.3461 0.3485 0.3508 0.3531 0.3554 0.3577 0.3599 0.36211.1 0.3643 0.3665 0.3686 0.3708 0.3729 0.3749 0.3770 0.3790 0.3810 0.38301.2 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3980 0.3997 0.40151.3 0.4032 0.4049 0.4066 0.4082 0.4099 0.4115 0.4131 0.4147 0.4162 0.41771.4 0.4192 0.4207 0.4222 0.4236 0.4251 0.4265 0.4279 0.4292 0.4306 0.43191.5 0.4332 0.4345 0.4357 0.4370 0.4382 0.4394 0.4406 0.4418 0.4429 0.44411.6 0.4452 0.4463 0.4474 0.4484 0.4495 0.4505 0.4515 0.4525 0.4535 0.45451.7 0.4554 0.4564 0.4573 0.4582 0.4591 0.4599 0.4608 0.4616 0.4625 0.46331.8 0.4641 0.4649 0.4656 0.4664 0.4671 0.4678 0.4686 0.4693 0.4699 0.47061.9 0.4713 0.4719 0.4726 0.4732 0.4738 0.4744 0.4750 0.4756 0.4761 0.47672.0 0.4772 0.4778 0.4783 0.4788 0.4793 0.4798 0.4803 0.4808 0.4812 0.48172.1 0.4821 0.4826 0.4830 0.4834 0.4838 0.4842 0.4846 0.4850 0.4854 0.48572.2 0.4861 0.4864 0.4868 0.4871 0.4875 0.4878 0.4881 0.4884 0.4887 0.48902.3 0.4893 0.4896 0.4898 0.4901 0.4904 0.4906 0.4909 0.4911 0.4913 0.49162.4 0.4918 0.4920 0.4922 0.4925 0.4927 0.4929 0.4931 0.4932 0.4934 0.49362.5 0.4938 0.4940 0.4941 0.4943 0.4945 0.4946 0.4948 0.4949 0.4951 0.49522.6 0.4953 0.4955 0.4956 0.4957 0.4959 0.4960 0.4961 0.4962 0.4963 0.49642.7 0.4965 0.4966 0.4967 0.4968 0.4969 0.4970 0.4971 0.4972 0.4973 0.49742.8 0.4974 0.4975 0.4976 0.4977 0.4977 0.4978 0.4979 0.4979 0.4980 0.49812.9 0.4981 0.4982 0.4982 0.4983 0.4984 0.4984 0.4985 0.4985 0.4986 0.49863.0 0.4987 0.4987 0.4987 0.4988 0.4988 0.4989 0.4989 0.4989 0.4990 0.4990

Areas Under the Normal CurveExample z = 1.96, then P(0 to z) = 0.4750

Table In Appendix Or On Inside Cover

33

Use The Standard Normal Probability Distribution To Find Area:

(Table On Inside Back Cover)

Computedz-score

Area undercurve

2.430.100.430.152.001.960.010.70

z 0 1.96

.475

35

Example 1 The daily water usage per person in New

Providence, New Jersey is normally distributed Mean = 20 gallons Standard deviation = 5 gallons

About 68% of those living in New Providence will use how many gallons of water?

+/- 1 standard deviation will give us: About 68% of the daily water usage will lie

between 15 and 25 gallons

36

Example 2 What is the probability that a person from

New Providence selected at random will use between 20 and 24 gallons per day?

00.05

2020

sX

z

80.05

2024

sX

z

37

Use The Table In The Back Of The Book And Look Up .80

The area under a normal curve between a z-value of 0 and a z-value of 0.80 is 0.2881

We conclude that 28.81 percent of the residents use between 20 and 24 gallons of water per day

See the following diagram:

38

- 5

0 . 4

0 . 3

0 . 2

0 . 1

. 0

x

f(

x

r a l i t r b u i o n : = 0 ,

-4 -3 -2 -1 0 1 2 3 4

z

Area =.2881“28.81% of the residents use between 20 and 24 gallons of water per day”

39

Example 3

What percent of the population use between 18 and 26 gallons per day?

40.05

2018

sX

z

20.15

2026

sX

z

40

The area associated with a z-value of –0.40 is .1554 Because the curve is symmetrical. look up .40 on

the right

The area associated with a z-value of 1.20 is .3849

.1554 + .3849 = .5403 We conclude that 54.03 percent of the

residents use between 18 and 26 gallons of water per day

Example 3

41

Example 4 Professor Mann has determined that the scores

in his statistics course are approximately normally distributed with a mean of 72 and a standard deviation of 5

He announces to the class that the top 15 percent of the scores will earn an A

What is the lowest score a student can earn and still receive an A?

.50 - .15 = .35 This is the area under the curve! You must look into table and find the value closest to .35

42

Table In Appendix Or On Inside Cover

z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.090.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239 0.0279 0.0319 0.03590.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.07530.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.0987 0.1026 0.1064 0.1103 0.11410.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.1368 0.1406 0.1443 0.1480 0.15170.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.1736 0.1772 0.1808 0.1844 0.18790.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088 0.2123 0.2157 0.2190 0.22240.6 0.2257 0.2291 0.2324 0.2357 0.2389 0.2422 0.2454 0.2486 0.2517 0.25490.7 0.2580 0.2611 0.2642 0.2673 0.2704 0.2734 0.2764 0.2794 0.2823 0.28520.8 0.2881 0.2910 0.2939 0.2967 0.2995 0.3023 0.3051 0.3078 0.3106 0.31330.9 0.3159 0.3186 0.3212 0.3238 0.3264 0.3289 0.3315 0.3340 0.3365 0.33891.0 0.3413 0.3438 0.3461 0.3485 0.3508 0.3531 0.3554 0.3577 0.3599 0.36211.1 0.3643 0.3665 0.3686 0.3708 0.3729 0.3749 0.3770 0.3790 0.3810 0.38301.2 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3980 0.3997 0.4015

Areas Under the Normal CurveExample z = 1.96, then P(0 to z) = 0.4750

43

Solve for X, The Score You Need To Get An A

The result is the score that separates students that earned an A from those that earned a B

Those with a score of 77.2 or more earn an A

721.04

51.04*5 72 77.2

X

X

44

Example 5

μ = 1000 = mean weekly income of shift foreman in the glass industryσ = 100

-3 -2 -1 0 1 2 3 zμ + -3 σ μ + -2 σ μ + -1 σ μ μ +1 σ μ +2 σ μ +3 σ x symbols

700 800 900 1000 1100 1200 1300 x in dollars

The weekly incomes of shift foremanin the glass industry are normally distributed

What is the probability of

selecting a shift foreman

whose salary isbetween $790

& $1200?

45

Example 5 Find Z-scores

790 10002.10

1001200 1000

2100

Xz

Xz

s

s

Look up area under the curve in the tables

46

Table In Appendix Or On Inside Cover

z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.090.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239 0.0279 0.0319 0.03590.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.07530.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.0987 0.1026 0.1064 0.1103 0.11410.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.1368 0.1406 0.1443 0.1480 0.15170.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.1736 0.1772 0.1808 0.1844 0.18790.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088 0.2123 0.2157 0.2190 0.22240.6 0.2257 0.2291 0.2324 0.2357 0.2389 0.2422 0.2454 0.2486 0.2517 0.25490.7 0.2580 0.2611 0.2642 0.2673 0.2704 0.2734 0.2764 0.2794 0.2823 0.28520.8 0.2881 0.2910 0.2939 0.2967 0.2995 0.3023 0.3051 0.3078 0.3106 0.31330.9 0.3159 0.3186 0.3212 0.3238 0.3264 0.3289 0.3315 0.3340 0.3365 0.33891.0 0.3413 0.3438 0.3461 0.3485 0.3508 0.3531 0.3554 0.3577 0.3599 0.36211.1 0.3643 0.3665 0.3686 0.3708 0.3729 0.3749 0.3770 0.3790 0.3810 0.38301.2 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3980 0.3997 0.40151.3 0.4032 0.4049 0.4066 0.4082 0.4099 0.4115 0.4131 0.4147 0.4162 0.41771.4 0.4192 0.4207 0.4222 0.4236 0.4251 0.4265 0.4279 0.4292 0.4306 0.43191.5 0.4332 0.4345 0.4357 0.4370 0.4382 0.4394 0.4406 0.4418 0.4429 0.44411.6 0.4452 0.4463 0.4474 0.4484 0.4495 0.4505 0.4515 0.4525 0.4535 0.45451.7 0.4554 0.4564 0.4573 0.4582 0.4591 0.4599 0.4608 0.4616 0.4625 0.46331.8 0.4641 0.4649 0.4656 0.4664 0.4671 0.4678 0.4686 0.4693 0.4699 0.47061.9 0.4713 0.4719 0.4726 0.4732 0.4738 0.4744 0.4750 0.4756 0.4761 0.47672.0 0.4772 0.4778 0.4783 0.4788 0.4793 0.4798 0.4803 0.4808 0.4812 0.48172.1 0.4821 0.4826 0.4830 0.4834 0.4838 0.4842 0.4846 0.4850 0.4854 0.48572.2 0.4861 0.4864 0.4868 0.4871 0.4875 0.4878 0.4881 0.4884 0.4887 0.4890

Areas Under the Normal CurveExample z = 1.96, then P(0 to z) = 0.4750

47

Example 5 The probability of selecting a shift foreman

whose salary is between $790 & $1200 is:

.4772 + .4821 = .9593

48

Example 6

μ = 1000 = mean weekly income of shift foreman in the glass industryσ = 100

-3 -2 -1 0 1 2 3 zμ + -3 σ μ + -2 σ μ + -1 σ μ μ +1 σ μ +2 σ μ +3 σ x symbols

700 800 900 1000 1100 1200 1300 x in dollars

The weekly incomes of shift foremanin the glass industry are normally distributed

What is the probability of

selecting a shift foreman

whose salary isless than $790?

49

Example 6

790 10002.10

100

Xz

s

Look up the areaThe area = .4821.5 - .4821 = .0179

The probability of selecting a shift foreman whose salary is less than $790 is .0179

50

51

Finding Area Under The Standard Normal Distribution – It’s All About Area!

-3 -2 -1 0 1 2 3 zμ + -3 σ μ + -2 σ μ + -1 σ μ μ +1 σ μ +2 σ μ +3 σ x symbols

1400 1600 1800 2000 2200 2400 2600 x in dollars

68.26% area

95.44% area

99.74% area

34.13% area

13.58% area

2.15% area

0.13% area

34.13% area

13.58% area

2.15% area

0.13% area

52

Finding Area Under The Standard Normal Distribution – Use Formulas & Tables

z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.090.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239 0.0279 0.0319 0.03590.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.07530.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.0987 0.1026 0.1064 0.1103 0.11410.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.1368 0.1406 0.1443 0.1480 0.15170.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.1736 0.1772 0.1808 0.1844 0.18790.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088 0.2123 0.2157 0.2190 0.22240.6 0.2257 0.2291 0.2324 0.2357 0.2389 0.2422 0.2454 0.2486 0.2517 0.25490.7 0.2580 0.2611 0.2642 0.2673 0.2704 0.2734 0.2764 0.2794 0.2823 0.2852

Areas Under the Normal CurveExample z = 1.96, then P(0 to z) = 0.4750

s

X

z

53

Finding Area Under The Standard Normal Distribution – Four Situations

1. If you wish to find the area between 0 and z (or – z), then you can look up the value directly in the table

2. If you wish to find the area beyond z (or –z), then locate the probability of z in the table and subtract it from .50

3. If you wish to find the area between two points on different sides of the mean, determine the z-values and add the corresponding areas

4. If you wish to find the area between two points on the same side of the mean, determine the z-values and subtract the smaller area from the larger

54

Use The Normal Distribution To Approximate The Binomial Probability Distribution

The normal distribution (a continuous distribution) yields a good approximation of the binomial distribution (a discrete distribution) for large values of n

Let’s Remember the conditions that must be met before we can run a binomial experiment

55

For An Experiment To Be Binomial It Must Satisfy The Following Conditions:

1. A random variable (x) counts the # of successes in a fixed number of trials (n)

Trials make up the experiment

2. Each trail must be independent of the previous trial

Outcome of one trial does not affect the outcome of any other trial

3. An outcome on each trial of an experiment is classified into one of two mutually exclusive categories:

1. Successor

2. Failure

4. The probability of success stays the same for each trial (so does the probability of failure)

56

Normal Approximating The Binomial

The normal probability distribution is generally a good approximation to the binomial probability distribution when:

n = Fixed # of trialsx = # of successes we wantp = "pi" = Probability of success of each trial

For Binomial Probability Distribution

n* > 5n*(1- > 5

Requirements for normal proabability distribution as an aproximation for a binomial probability distribution

57

Mean & Variance of the Binomial Distribution

n

n = Fixed # of trialsp = Probability of success of each trial

μ = Mean of binomial distributionσ = Standard deviation of binomial distribution

For Binomial Probability Distribution

2 (1 )ns s Empirical experiments have shown these to be acceptable estimates

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Continuity Correction Factor The continuity correction factor of .5 is used

to extend the continuous value of x one-half unit in either direction

The correction compensates for estimating a discrete distribution by a continuous distribution

When you use discrete numbers, you have “gaps” – you need to take an average that will yield a number between

We will simply estimate by adding or subtracting the value .5

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Continuity Correction Factor1. For the probability at least x occur, use the

area above (x - .5)

2. For the probability that more than x occur, use the area above (x + .5)

3. For the probability that x or fewer occur, use the area below (x + .5)

4. For the probability that fewer than x occur, use the area below (x - .5)

Example Example

A recent study by a marketing research firm showed that 15% of American households owned a video camera.

For a sample of 200 homes, What is the probability that less than 40 homes in the sample have video cameras?

Step 1: Binomial?

1. Fixed Trails = yes, n = 200

2. Independent = yes

3. S/F Success = have video camera, Failure = don’t have

4. Constant = .15

Example Example

Step 2 : Can we approximate binomial distribution with a standard normal distribution?

(.15)(200) 30 5

(1 ) (.85)(200) 170 5

n

n

Step 3 : Calculate μ & σ for the binomial distribution

n (. )( )15 200 302 (1 ) (30)(1 .15) 5.0498ns

Step 4: Calculate z-value (.5 correction)Step 4: Calculate z-value (.5 correction)

What is the probability that less than 40 homes in the sample have video cameras?

We use the correction factor, so X is 39.5

The value of z is 1.88

88.10498.5

0.305.39

sX

z

Step 5: Look Up AreaStep 5: Look Up Area

From Appendix the area between 0 and 1.88 on the z scale is .4699

So the area to the left of 1.88 is .5000 + .4699 = .9699

The likelihood that less than 40 of the 200 homes have a video camera is about 97%

- 5

0 . 4

0 . 3

0 . 2

0 . 1

. 0

f(

x

r a l i t r b u i o n : = 0 , s = 1

EXAMPLE 5

0 1 2 3 4

Area = .5000+.4699 =.9699

z=1.88

z

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Summarize Chapter 7

1. Understand the difference between discrete and continuous distributions

2. Compute the mean and the standard deviation for a uniform distribution

3. Compute probabilities using the uniform distribution

4. List the characteristics of the:• Normal probability distribution• Standard normal probability distribution

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Summarize Chapter 7

1. Define and calculate z values2. Use the standard normal probability

distribution to find area: Above the mean Below the mean Between two values Above one value Below one value

3. Use the normal distribution to approximate the binomial probability distribution